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A350332
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Numbers p^2*q, p < q odd primes such that p does not divide q-1.
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5
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45, 99, 153, 175, 207, 261, 325, 369, 423, 425, 475, 477, 531, 539, 575, 637, 639, 725, 747, 801, 833, 909, 925, 931, 963, 1017, 1075, 1127, 1175, 1179, 1233, 1325, 1341, 1475, 1503, 1519, 1557, 1573, 1611, 1675, 1719, 1773, 1813, 1825, 1975, 2009, 2043, 2057
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OFFSET
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1,1
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COMMENTS
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For these terms m, there are precisely 2 groups of order m, so this is a subsequence of A054395.
The 2 groups are abelian; they are C_{p^2*q} and (C_p X C_p) X C_q, where C means cyclic groups of the stated order and the symbol X means direct product.
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REFERENCES
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Pascal Ortiz, Exercices d'Algèbre, Collection CAPES / Agrégation, Ellipses, problème 1.35, pp. 70-74, 2004.
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LINKS
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EXAMPLE
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99 = 3^2 * 11, 3 and 11 are odd and 3 does not divide 11-1 = 10, hence 99 is a term.
175 = 5^2 * 7, 5 and 7 are odd and 5 does not divide 7-1 = 6, hence 115 is another term.
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MATHEMATICA
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q[n_] := Module[{f = FactorInteger[n], p, e}, p = f[[;; , 1]]; e = f[[;; , 2]]; e == {2, 1} && ! Divisible[p[[2]] - 1, p[[1]]]]; Select[Range[2000], q] (* Amiram Eldar, Dec 25 2021 *)
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PROG
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(Python)
from sympy import integer_nthroot, primerange
def aupto(limit):
aset, maxp = set(), integer_nthroot(limit, 3)[0]
for p in primerange(3, maxp+1):
pp = p*p
for q in primerange(p+1, limit//pp+1):
if (q-1)%p != 0:
aset.add(pp*q)
return sorted(aset)
(PARI) isok(m) = my(f=factor(m)); if (f[, 2] == [2, 1]~, my(p=f[1, 1], q=f[2, 1]); ((q-1) % p)); \\ Michel Marcus, Dec 25 2021
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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